Răspuns:
c
Explicație pas cu pas:
[tex]\displaystyle 2C_{n+2}^3 = 2\cdot \frac{(n+2)!}{3!(n+2-3)!} = \frac{\not{2}(n+2)!}{1\cdot \not{2}\cdot 3 \cdot (n-1)!} = \frac{\not{1}\cdot \not{2}\cdot \not{3}\cdot \cdots \cdot \not{(n-1)}\cdot n \cdot (n+1)\cdot (n+2)}{3\cdot \not{1}\cdot \not{2}\cdot \not{3}\cdot \cdots \cdot \not{(n-1)}} = \frac{n(n+1)(n+2)}{3}\\\\\sum_{k=1}^{n} k(ak+b) = \sum_{k=1}^{n} (ak^2 + bk) = \sum_{k=1}^{n}ak^2 + \sum_{k=1}^{n}bk = a\Bigg(\sum_{k=1}^{n}k^2\Bigg) + b\Bigg(\sum_{k=1}^{n} k\Bigg) = a\cdot \frac{n(n+1)(2n+1)}{6} + b\cdot \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}\Bigg|\div n(n+1)\\\\ a\cdot \frac{2n+1}{6} + b\cdot \frac{1}{2} = \frac{n+2}{3}\Bigg|\cdot 6\\\\ a(2n+1) + 3b = 2(n+2)\\\\\textrm{Acum daca ii dam lui n valoarea 1}(1 \in \mathbb{N}^*)\\\\\implies a\cdot (2\cdot 1 + 1) + 3b = 2\cdot (1+2)\\\\ 3a + 3b = 2\cdot 3\\\\ 3(a+b) = 2\cdot 3\\\\ \boxed{a+b = 2} (1)\\\\\textrm{Daca ii dam lui n valoarea 2:} \\\\a\cdot (2\cdot 2 + 1) + 3b = 2\cdot (2+2) \\\\ 5a + 3b = 2\cdot 4\\\\ 3a + 3b + 2a = 8\\\\ 3(a+b) + 2a = 8\\\\ 3\cdot 2 + 2a = 8\\\\ 6+2a = 8\\\\ 2a = 8 - 6 = 2\\\\ \bodxed{a = 1} \\\\ \implies b = 2 - 1 \implies \boxed{b = 1}[/tex]
